Voltage drop across a wire/trace- return path confusion

[dany.1986]

Hi

I am trying to figure out what would be the voltage seen by a load after the voltage drop in a wire/trace. I know that to calculate the voltage drop in a wire/trace we simply use the Ohm's law and then substract it from the voltage source to get the final voltage seen by the load. What I am confused about is why do we have to include the return path as well. In example, if Vsource=10 V, voltage drop of a source wire= 2 V then the voltage seen by the load would be 8 V, however, if we include the return wire which has the voltage drop of 2 V then the load would get 6 V. Why do we have to include the return path? Is it going to be the same case with the load directly on a PCB but with a long return trace to ground?

Thank you for your help.

 

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If there is current flowing in a path:
1) from the source
2) thru source wire
3) thru load
4) thru return wire
5) back to the source

Then there will be voltage drops at points 2 and 4 due to wire resistance. The resistance is typically very small in most applications so there is hardly any voltage drop, unless there are very high currents involved, e.g. FPGA core voltages of 1.2V may draw 4A so even a 1" 0.16 ohm trace would result in a 0.64V drop.

That is why you don't want to supply an FPGA voltage rail with a 6 mil trace using 0.5 oz copper.

 

[dany.1986]

So it means that if the return trace to ground is also 1" 0.16 ohm then the total voltage drop will be 1.28 V?

 

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Yes, but it also means nothing will work as you can't both be having a 1.28 V voltage drop on a 1.2V supply, in this theoretical situation you'll probably see the FPGA is starved for current and doesn't operate correctly and the current draw on the 1.2V supply is much lower than the 4A.